An ideal must also be a subset of R which I=<a>U<b> is so I don't see a problem with I being an ideal.

Then you''re not thinking very hard. Or not doing some examples. Or making a mistake with what <a>U<b> is. Something is in <a>U<b> if it is <a> OR it is in <b>. That will not in general make an ideal. In fact it is an ideal if and only if <a> is contained in <b> or vice versa.

One thing I've noticed, pixova, is that you don't seem to work out any examples to figure out what is going on. Intuition is good, but you need to develop it, and examples help.

Let's take the simplest example of Z, <2> and <3>. What is <2>U<3>? It is the set of things that are multiples of 2 *or* of 3. That is not an ideal.

Is that a typo? 10 is certainly in <2>U<3>, since it is a multiple of 2.

On the example thing, if 10 wasn't a typo, the example is yours to play with, so you can always take the simplest thing - I'm always amazed at students who come up with some weird or wonderful counter example when frequently, say, 0 would do.

Of course here 0 isn't the right thing to look at. But what about 1? Or what about 2+3? It is the integers we're talking about here. The integers are PID. You've known for ages (the euclidean algorithm) that it is a PID, even if you don't know the words: <a,b>=<gcd(a,b)>, so picking any two coprime elements like 2 and 3, <2,3>=Z=/=<2>U<3>.

So, using knowledge of Z, and the good old Euclidean algorithm tells you it is wrong in Z.

Now, by picking the *simplest* possible thing to look at show that <a>U<b> is an ideal in any ring if and only if <a> is a subset of <b> or vice versa.

Mybad, an error that went unnoticed by my behalf. 1 would be a counter example. I have a more general question:

When we talk about <a,b> we have to mention whether it is generating a subring or an ideal. If an ideal than <a,b> = {Aa+Bb|all A,B in R} as this set is the smallest ideal containing both a and b.

However what about <a,b> as generating a subring of R? It would have to be the smallest subring containing a and b. However it may be a smaller set than {Aa+Bb|all A,B in R} although in some cases it is exactly that set. i.e <2,x> as an ideal or subring of Z[x] is the set of all polynomials in Z[x] with even constant term, including 0. Is there a formula or algorithm to determine <a,b> as generating a subring of R?

From what I have studied, 1 dosen't need to be in a ring. But 0 must be.

My lecturer talks about cases of a ring containing 1 which implicitly means there are rings without 1 but we are talking about rings with 1. He would never 'say' a ring with 0 because it would be a tautology as 0 must by definition be in any ring.

Is [itex]\mathbb{Q}(\sqrt{2}) \subseteq \mathbb{C}[/itex] just {a+bsqrt(2)|a,b in Q}

From what I have studied, 1 dosen't need to be in a ring. But 0 must be.

My lecturer talks about cases of a ring containing 1 which implicitly means there are rings without 1 but we are talking about rings with 1. He would never 'say' a ring with 0 because it would be a tautology as 0 must by definition be in any ring.

Is [itex]\mathbb{Q}(\sqrt{2}) \subseteq \mathbb{C}[/itex] just {a+bsqrt(2)|a,b in Q}

I always hear rumors that there are people out there who define the word "ring" without requiring that it have 1, but I've never actually met someone or read a text that does so.

I've only ever seen "ring with 1" used as a synonym for "ring", to clarify the situation if the reader/listener wasn't sure what definition the author/speaker is using. I've only ever seen "ring without 1" used as an entire phrase -- i.e. "ring without 1" does not mean "ring" "without 1". In particular a "ring without 1" need not be a ring, and may, in fact, have a multiplicative identity.

Here is something new for you, "A more significant difference is that some authors (such as I. N. Herstein) omit the requirement that a ring have a multiplicative identity. These authors call rings which do have multiplicative identities unital rings, unitary rings, or simply rings with unity." in http://en.wikipedia.org/wiki/Ring_(mathematics)

Really? Perhaps, because every ring can be extended to a unital ring, and there is a universal such, however, that extension might ruin some important properties. To be honest, the only non-unital ring that I've met was the subspace of Endos of a countable dimensional banach space with finite trace.